Question 1207461
I made a mistake on my first solution. Here is the correct version:
<pre>
The values of the coins are irrelevant. Also, the probabilities of
selecting coins is also irrelevant.  [In future problems like this
they will be relevant].  

But all that matters is that you started with 6 coins, two pairs 
and two singles, where each member of each pair is indistinguishable 
from the other member of that pair.  Say, the 6 coins are AABBCD.

Case 1. You choose one of the A's not to replace. You have left ABBCD.
You can then choose AB,AC,AD,BB,BC,BD,CD
There are 7 outcomes.  

Case 2. You choose one of the B's not to replaced.  Same as case 1. 
There are 7 outcomes.

Case 3. You choose the C not to replace. You have left AABBD.
You can then choose AA,AB,AD,BB,BD
There are 5 outcomes.

Case 4. You choose the D not to replace.  Same as case 3. 
There are 5 outcomes.

Total = 7+7+5+5=22 possible outcomes.

Edwin</pre>